C how can I decrease my program size - c

I am writing a c program for a uController and I managed to fill 32kb with code.
The program is full with the following 'print' function:
void print(char *x)
{
while(*x) {
SerialWrite(*x++);}
}
SerialWrite looks like:
void SerialWrite(unsigned char c)
{
while(tx_buffer_size>250);
ES=0;
tx_buffer_size++;
if (tx_buffer_empty == 0){
txBuffer[tx_in++] = c;}
else {
tx_buffer_empty = 0;
SBUF = c;}
ES=1;
}
I call the print function with a string text as arguement like:
print("Hello World");
I only used 2.2kb of my 32kb external memory so my conclusion is too move "Hello World" to the external memory instead of having it hardcoded in the main program.
Unfortunately my attempts actually made it worse, I did the following:
char xdata *msg1 = "Hello World"; // <-- this made it worse, and the external memory space was not used at all
print(msg1);
Than I tried:
char xdata msg1[] = "Hello World";
print(msg1);
This did increase the external memory size by 12 which is correct as the string contains 11 chars + the null. But the program memory also increased by 3. I tried different string lengths but the program memory keeps increasing by 3 bytes.
How can I solve this problem?
Addition:
I am using Keil's C compiler and I compile with favor for program size.
The uController is an FPGA chip which has an emulated 80C51 chip. For this virtual 80c51 I am writing the code. I have 32kb memory for the main program and 32kb for variables
EDIT: *msg[] was a typo, it had to be *msg

char xdata *msg1 = "Hello World";
char xdata msg1[] = "Hello World";
I don't think you can ever shorten your program this way, because even if the message will be accessed from xdata, it will be first copied there by the C runtime, fetching it from code (initialized data).
If you really have a lot of "print()", then you can squeeze some bytes by declaring a correct pointer type for its parameter. I explain: due to the strange architecture of the 8051, a "high level" C "generic" pointer takes three bytes - two bytes as offset, and a third byte to specify whether it points in XDATA or in CODE. Then every call to print() has to construct such a pointer.
If, for example, you declare:
void print(char xdata *x)
now print() only accepts pointers into xdata, and only two bytes long pointers are required. Every call to print() will be simpler.
It seems that most of your texts will be constant, so they will be in code; and so the pointer type you need is not XDATA but CODE (or what else keyword is correct).
Once you have declared print() like above, you will NO MORE be able to print() data which is not in the code segment; perhaps you will need another function, like
printx(char* what)
which will accept char* from any segments, like before, but you will call this costly printx() only when needed.
Hope this helps; the 8051 is very peculiar and compilers for it are complicated (but Keil is superb). I would suggest to take a look at the generated (low level) code, may be you can spot some other memory waste. Good luck!
Addendum:______________
It happened once to me to have little room for texts. I solved nicely by applying some text compression. Given that all the texts were plain ASCII, I used characters over 127 to point to common, repeated words in the texts. For example "Hello world" can be replaced with 0x80 " world", if the string "Hello" is repeated a few times. Of course, the print routine must detect those special markers and manage the decompression - but this is very easy.

In short, put this line in the beginning of your source file:
#pragma STRING (XDATA)
These pages may help you understand better:
http://www.keil.com/support/man/docs/c51/c51_le_const.htm
http://www.keil.com/support/man/docs/c51/c51_string.htm
2018/06/25 edit:
try this line:
#pragma O2 STRING (XDATA)
as stated in the 2nd link above and this page, OMF2 should be enabled to use STRING directive
http://www.keil.com/support/man/docs/c51/c51_omf2.htm

Related

C18 Passing a char array to function

I am new to C programming and microcontrollers. I am using a PIC18F24K20 microcontroller with C18. I have it set up to receive information from a computer input using USART transmit and receive functions. My goal is to compare the received word against known words, and transmit something back to the computer based on what word was received. Below is the relevant code.
#include "p18f24k20.h"
#include "delays.h"
#include "string.h"
#include "stdlib.h"
void CommTransmit ( rom char * );
void main (void)
{
char buf[11], data, T;
int i;
i = 0;
memset(buf, 0, sizeof buf);
while(1)
{
if (PIR1bits.RCIF)
{
data = USART_receive();
if (data != 47) // 47 is /, indicates end of string
{
buf[i] = data;
i++;
}
else
{
// T = strcmppgm2ram(buf,(const far rom char*)"test");
CommTransmit(buf);
USART_transmit('t');
buf[0] = 0'
}
}
}
}
void CommTransmit ( rom char *CommVariable )
{
char test;
test = strcmppgm2ram(CommVariable, (const far rom char*)"test");
if (test == 0)
{
USART_transmit('g');
}
}
The code is currently set up to test to try to determine what is wrong. If I run it as is, the computer will receive a 't', as if the microcontroller ran through the CommTransmit function. However, it never transmits the 'g'. Even if I put a USART_transmit('g') call in the CommTransmit function, outside of and after the if statement, it never gets called (like it gets stuck in the strcmppgm2ram function?) but yet it still transmits the 't'.
It is also strange because if I put a break at the CommTransmit function and run through line by line, it seems to work properly. However, if I watch the CommVariable inside MPLAB IDE, it is never what it supposed to be (though the 'buf' variable prior to being called into the function is correct). From what I can tell, the value of CommVariable when I watch it depends on the size of the array.
From reading, I think it may caused by how the microcontroller stores the variable (program vs data memory?) but I'm not sure. Any help is greatly appreciated!
edit: I should also add that if I uncomment the T = strcmppgm2ram line in the else statement before the CommTransmit line, it works properly (T = 0 when the two strings are the same). I believe the array changes when I pass it through the function, which causes the strcmppgm2ram function to not work properly.
Looking at signature for strcmppgm2ram
signed char strcmppgm2ram(const char * str1, const rom char * str2 );
I don't understand why do you have rom char * for CommVariable. From chapter 2.4.3 ram/rom Qualifiers of MPLAB® C18 C Compiler User’s Guide
Because the PICmicro microcontrollers use separate program memory and
data memory address busses in their design, MPLAB C18 requires
extensions to distinguish between data located in program memory and
data located in data memory. /---/ Pointers can point to either data memory (ram pointers) or program
memory (rom pointers). Pointers are assumed to be ram pointers unless
declared as rom.
And in 2.7.3 String Constants:
An important consequence of the separate address spaces for MPLAB C18
is that pointers to data in program memory and pointers to data in
data memory are not compatible. /---/ because they refer to different
address spaces. /---/
MPLAB C18 automatically places all string constants in program memory.
This type of a string constant is “array of char located in program
memory”, (const rom char []).
And also it's not clear the purpose of type casting to const far rom char* for the second argument. That may cause stack corruption because far pointer has bigger size (24 bits). So, it looks like it should be rewritten as:
void CommTransmit (const char *CommVariable )
{
if (!strcmppgm2ram(CommVariable, "test")) {
USART_transmit('g');
}
}

Why does my program throw a segmentation fault while using heap-allocated memory?

After writing a program to reverse a string, I am having trouble understanding why I got a seg fault while trying to reverse the string. I have listed my program below.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void reverse(char *);
int main() {
char *str = calloc(1,'\0');
strcpy(str,"mystring0123456789");
reverse(str);
printf("Reverse String is: %s\n",str);
return 0;
}
void reverse(char *string) {
char ch, *start, *end;
int c=0;
int length = strlen(string);
start = string;
end = string;
while (c < length-1){
end++;
c++;
}
c=0;
while(c < length/2){
ch = *end;
*end = *start;
*start = ch;
start++;
end--;
c++;
}
}
1st Question:
Even though I have allocated only 1 byte of memory to the char pointer
str (calloc(1,'\0')), and I copied a 18 bytes string mystring0123456789 into it, and it didn't throw any error and the program worked fine without any SEGFAULT.
Why did my program not throw an error? Ideally it should throw some error as it don't have any memory to store that big string. Can someone throw light on this?
The program ran perfectly and gives me output Reverse String is: 9876543210gnirtsym.
2nd Question:
If the replace the statement
strcpy(str,"mystring0123456789");
with
str="mystring0123456789\0";
the program gives segmentation fault even though I have allocated enough memory for str (malloc(100)).
Why the program throwing segmentation fault?
Even though i have allocated only 1 byte of memory to the char pointer str(calloc(1,'\0')), and i copied a 18 bytes string "mystring0123456789" into it, and it didn't throw any error and the program worked fine without any SEGFAULT.
Your code had a bug -- of course it's not going to do what you expect. Fix the bug and the mystery will go away.
If the replace the statement
strcpy(str,"mystring0123456789");
with
str="mystring0123456789\0";
the program gives segmentation fault even though i have allocated enough memory for str (malloc(100)).
Because when you finish this, str points to a constant. This throws away the previous value of str, a pointer to memory you allocated, and replaces it with a pointer to that constant.
You cannot modify a constant, that's what makes it a constant. The strcpy function copies the constant into a variable which you can then modify.
Imagine if you could do this:
int* h = &2;
Now, if you did *h = 1; you'd be trying to change that constant 2 in your code, which of course you can't do.
That's effectively what you're doing with str="mystring0123456789\0";. It makes str point to that constant in your source code which, of course, you can't modify.
There's no requirement that it throw a segmentation fault. All that happens is that your broken code invokes undefined behavior. If that behavior has no visible effect, that's fine. If it formats the hard drive and paints the screen blue, that's fine too. It's undefined.
You're overwriting the pointer value with the address of a string literal, which totally doesn't use the allocated memory. Then you try to reverse the string literal which is in read-only memory, which causes the segmentation fault.
Your program did not throw an error because, even though you did the wrong thing, ncaught you (more below). You wrote data were you were not supposed to, but you got “lucky” and did not break anything by doing this.
strcpy(str,"mystring0123456789"); copies data into the place where str points. It so happens that, at that place, you are able to write data without causing a trap (this time). In contrast, str="mystring0123456789\0"; changes str to point to a new place. The place it points to is the place where "mystring0123456789\0" is stored. That place is likely read-only memory, so, when you try to write to it in the reverse routine, you get a trap.
More about 1:
When calloc allocates memory, it merely arranges for there to be some space that you are allowed to use. Physically, there is other memory present. You can write to that other memory, but you should not. This is exactly the way things work in the real world: If you rent a hotel room, you are allowed to use that hotel room, but it is wrong for you to use other rooms even if they happen to be open.
Sometimes when you trespass where you are not supposed to, in the real world or in a program, nobody will see, and you will get away with it. Sometimes you will get caught. The fact that you do not get caught does not mean it was okay.
One more note about calloc: You asked it to allocate space for one thing of zero size (the source code '\0' evaluates to zero). So you are asking for zero bytes. Various standards (such as C and Open Unix) may say different things about this, so it may be that, when you ask for zero bytes, calloc gives you one byte. However, it certainly does not give you as many bytes as you wrote with strcpy.
It sounds like you are writing C programs having come from a dynamic language or at least a language that does automatic string handling. For lack of a more formal definition, I find C to be a language very close to the architecture of the machine. That is, you make a lot of the programming decisions. A lot of your program problems are the result of your code causing undefined behavior.You got a segfault with strcpy, because you copied memory into a protected location; the behavior was undefined. Whereas, assigning your fixed string "mystring0123456789\0" was just assigning that pointer to str.
When you implement in C, you decide whether you want to define your storage areas at compile or run-time, or decide to have storage allocated from the heap (malloc/calloc). In either case, you have to write housekeeping routines to make sure you do not exceed the storage you have defined.
Assigning a string to a pointer merely assigns the string's address in memory; it does not copy the string, and a fixed string inside quotes "test-string" is read-only, and you cannot modify it. Your program may have worked just fine, having done that assignment, even though it would not be considered good C coding practice.
There are advantages to handling storage allocations this way, which is why C is a popular language.
Another case is that you can have a segfault when you use memory correct AND your heap became so big that your physical memory cannot manage it (without overlap with stack|text|data|bss -> link)
Proof: link, section Possible Cause #2

How strcpy works behind the scenes?

This may be a very basic question for some. I was trying to understand how strcpy works actually behind the scenes. for example, in this code
#include <stdio.h>
#include <string.h>
int main ()
{
char s[6] = "Hello";
char a[20] = "world isnsadsdas";
strcpy(s,a);
printf("%s\n",s);
printf("%d\n", sizeof(s));
return 0;
}
As I am declaring s to be a static array with size less than that of source. I thought it wont print the whole word, but it did print world isnsadsdas .. So, I thought that this strcpy function might be allocating new size if destination is less than the source. But now, when I check sizeof(s), it is still 6, but it is printing out more than that. Hows that working actually?
You've just caused undefined behaviour, so anything can happen. In your case, you're getting lucky and it's not crashing, but you shouldn't rely on that happening. Here's a simplified strcpy implementation (but it's not too far off from many real ones):
char *strcpy(char *d, const char *s)
{
char *saved = d;
while (*s)
{
*d++ = *s++;
}
*d = 0;
return saved;
}
sizeof is just returning you the size of your array from compile time. If you use strlen, I think you'll see what you expect. But as I mentioned above, relying on undefined behaviour is a bad idea.
http://natashenka.ca/wp-content/uploads/2014/01/strcpy8x11.png
strcpy is considered dangerous for reasons like the one you are demonstrating. The two buffers you created are local variables stored in the stack frame of the function. Here is roughly what the stack frame looks like:
http://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Call_stack_layout.svg/342px-Call_stack_layout.svg.png
FYI things are put on top of the stack meaning it grows backwards through memory (This does not mean the variables in memory are read backwards, just that newer ones are put 'behind' older ones). So that means if you write far enough into the locals section of your function's stack frame, you will write forward over every other stack variable after the variable you are copying to and break into other sections, and eventually overwrite the return pointer. The result is that if you are clever, you have full control of where the function returns. You could make it do anything really, but it isn't YOU that is the concern.
As you seem to know by making your first buffer 6 chars long for a 5 character string, C strings end in a null byte \x00. The strcpy function copies bytes until the source byte is 0, but it does not check that the destination is that long, which is why it can copy over the boundary of the array. This is also why your print is reading the buffer past its size, it reads till \x00. Interestingly, the strcpy may have written into the data of s depending on the order the compiler gave it in the stack, so a fun exercise could be to also print a and see if you get something like 'snsadsdas', but I can't be sure what it would look like even if it is polluting s because there are sometimes bytes in between the stack entries for various reasons).
If this buffer holds say, a password to check in code with a hashing function, and you copy it to a buffer in the stack from wherever you get it (a network packet if a server, or a text box, etc) you very well may copy more data from the source than the destination buffer can hold and give return control of your program to whatever user was able to send a packet to you or try a password. They just have to type the right number of characters, and then the correct characters that represent an address to somewhere in ram to jump to.
You can use strcpy if you check the bounds and maybe trim the source string, but it is considered bad practice. There are more modern functions that take a max length like http://www.cplusplus.com/reference/cstring/strncpy/
Oh and lastly, this is all called a buffer overflow. Some compilers add a nice little blob of bytes randomly chosen by the OS before and after every stack entry. After every copy the OS checks these bytes against its copy and terminates the program if they differ. This solves a lot of security problems, but it is still possible to copy bytes far enough into the stack to overwrite the pointer to the function to handle what happens when those bytes have been changed thus letting you do the same thing. It just becomes a lot harder to do right.
In C there is no bounds checking of arrays, its a trade off in order to have better performance at the risk of shooting yourself in the foot.
strcpy() doesn't care whether the target buffer is big enough so copying too many bytes will cause undefined behavior.
that is one of the reasons that a new version of strcpy were introduced where you can specify the target buffer size strcpy_s()
Note that sizeof(s) is determined at run time. Use strlen() to find the number of characters s occupied. When you perform strcpy() source string will be replaced by destination string so your output wont be "Helloworld isnsadsdas"
#include <stdio.h>
#include <string.h>
main ()
{
char s[6] = "Hello";
char a[20] = "world isnsadsdas";
strcpy(s,a);
printf("%s\n",s);
printf("%d\n", strlen(s));
}
You are relying on undefined behaviour in as much as that the compiler has chose to place the two arrays where your code happens to work. This may not work in future.
As to the sizeof operator, this is figured out at compile time.
Once you use adequate array sizes you need to use strlen to fetch the length of the strings.
The best way to understand how strcpy works behind the scene is...reading its source code!
You can read the source for GLibC : http://fossies.org/dox/glibc-2.17/strcpy_8c_source.html . I hope it helps!
At the end of every string/character array there is a null terminator character '\0' which marks the end of the string/character array.
strcpy() preforms its task until it sees the '\0' character.
printf() also preforms its task until it sees the '\0' character.
sizeof() on the other hand is not interested in the content of the array, only its allocated size (how big it is supposed to be), thus not taking into consideration where the string/character array actually ends (how big it actually is).
As opposed to sizeof(), there is strlen() that is interested in how long the string actually is (not how long it was supposed to be) and thus counts the number of characters until it reaches the end ('\0' character) where it stops (it doesn't include the '\0' character).
Better Solution is
char *strcpy(char *p,char const *q)
{
char *saved=p;
while(*p++=*q++);
return saved;
}

C char* pointers pointing to same location where they definitely shouldn't

I'm trying to write a simple C program on Ubuntu using Eclipse CDT (yes, I'm more comfortable with an IDE and I'm used to Eclipse from Java development), and I'm stuck with something weird. On one part of my code, I initialize a char array in a function, and it is by default pointing to the same location with one of the inputs, which has nothing to do with that char array. Here is my code:
char* subdir(const char input[], const char dir[]){
[*] int totallen = strlen(input) + strlen(dir) + 2;
char retval[totallen];
strcpy(retval, input);
strcat(retval, dir);
...}
Ok at the part I've marked with [*], there is a checkpoint. Even at that breakpoint, when I check y locals, I see that retval is pointing to the same address with my argument input. It not even possible as input comes from another function and retval is created in this function. Is is me being unexperienced with C and missing something, or is there a bug somewhere with the C compiler?
It seems so obvious to me that they should't point to the same (and a valid, of course, they aren't NULL) location. When the code goes on, it literally messes up everything; I get random characters and shapes in console and the program crashes.
I don't think it makes sense to check the address of retval BEFORE it appears, it being a VLA and all (by definition the compiler and the debugger don't know much about it, it's generated at runtime on the stack).
Try checking its address after its point of definition.
EDIT
I just read the "I get random characters and shapes in console". It's obvious now that you are returning the VLA and expecting things to work.
A VLA is only valid inside the block where it was defined. Using it outside is undefined behavior and thus very dangerous. Even if the size were constant, it still wouldn't be valid to return it from the function. In this case you most definitely want to malloc the memory.
What cnicutar said.
I hate people who do this, so I hate me ... but ... Arrays of non-const size are a C99 extension and not supported by C++. Of course GCC has extensions to make it happen.
Under the covers you are essentially doing an _alloca, so your odds of blowing out the stack are proportional to who has access to abuse the function.
Finally, I hope it doesn't actually get returned, because that would be returning a pointer to a stack allocated array, which would be your real problem since that array is gone as of the point of return.
In C++ you would typically use a string class.
In C you would either pass a pointer and length in as parameters, or a pointer to a pointer (or return a pointer) and specify the calls should call free() on it when done. These solutions all suck because they are error prone to leaks or truncation or overflow. :/
Well, your fundamental problem is that you are returning a pointer to the stack allocated VLA. You can't do that. Pointers to local variables are only valid inside the scope of the function that declares them. Your code results in Undefined Behaviour.
At least I am assuming that somewhere in the ..... in the real code is the line return retval.
You'll need to use heap allocation, or pass a suitably sized buffer to the function.
As well as that, you only need +1 rather than +2 in the length calculation - there is only one null-terminator.
Try changing retval to a character pointer and allocating your buffer using malloc().
Pass the two string arguments as, char * or const char *
Rather than returning char *, you should just pass another parameter with a string pointer that you already malloc'd space for.
Return bool or int describing what happened in the function, and use the parameter you passed to store the result.
Lastly don't forget to free the memory since you're having to malloc space for the string on the heap...
//retstr is not a const like the other two
bool subdir(const char *input, const char *dir,char *retstr){
strcpy(retstr, input);
strcat(retstr, dir);
return 1;
}
int main()
{
char h[]="Hello ";
char w[]="World!";
char *greet=(char*)malloc(strlen(h)+strlen(w)+1); //Size of the result plus room for the terminator!
subdir(h,w,greet);
printf("%s",greet);
return 1;
}
This will print: "Hello World!" added together by your function.
Also when you're creating a string on the fly you must malloc. The compiler doesn't know how long the two other strings are going to be, thus using char greet[totallen]; shouldn't work.

C Duration of strings, constants, compound literals, and why not, the code itself

I didn't remember where I read, that If I pass a string to a function like.
char *string;
string = func ("heyapple!");
char *func (char *string) {
char *p
p = string;
return p;
}
printf ("%s\n", string);
The string pointer continue to be valid because the "heyapple!" is in memory, it IS in the code the I wrote, so it never will be take off, right?
And about constants like 1, 2.10, 'a'?
And compound literals?
like If I do it:
func (1, 'a', "string");
Only the string will be all of my program execution, or the constans will be too?
For example I learned that I can take the address of string doing it
&"string";
Can I take the address of the constants literals? like 1, 2.10, 'a'?
I'm passing theses to functions arguments and it need to have static duration like strings without the word static.
Thanks a lot.
This doesn't make a whole lot of sense.
Values that are not pointers cannot be "freed", they are values, they can't go away.
If I do:
int c = 1;
The variable 'c' is not a pointer, it cannot do anything else than contain an integer value, to be more specific it can't NOT contain an integer value. That's all it does, there are no alternatives.
In practice, the literals will be compiled into the generated machine-code, so that somewhere in the code resulting from the above will be something like
load r0, 1
Or whatever the assembler for the underlying instruction set looks like. The '1' is a part of the instruction encoding, it can't go away.
Make sure you distinguish between values and pointers to memory. Pointers are themselves values, but a special kind of value that contains an address to memory.
With char* hello = "hello";, there are two things happening:
the string "hello" and a null-terminator are written somewhere in memory
a variable named hello contains a value which is the address to that memory
With int i = 0; only one thing happens:
a variable named i contains the value 0
When you pass around variables to functions their values are always copied. This is called pass by value and works fine for primitive types like int, double, etc. With pointers this is tricky because only the address is copied; you have to make sure that the contents of that address remain valid.
Short answer: yes. 1 and 'a' stick around due to pass by value semantics and "hello" sticks around due to string literal allocation.
Stuff like 1, 'a', and "heyapple!" are called literals, and they get stored in the compiled code, and in memory for when they have to be used. If they remain or not in memory for the duration of the program depends on where they are declared in the program, their size, and the compiler's characteristics, but you can generally assume that yes, they are stored somewhere in memory, and that they don't go away.
Note that, depending on the compiler and OS, it may be possible to change the value of literals, inadvertently or purposely. Many systems store literals in read-only areas (CONST sections) of memory to avoid nasty and hard-to-debug accidents.
For literals that fit into a memory word, like ints and chars it doesn't matter how they are stored: one repeats the literal throughout the code and lets the compiler decide how to make it available. For larger literals, like strings and structures, it would be bad practice to repeat, so a reference should be kept.
Note that if you use macros (#define HELLO "Hello!") it is up to the compiler to decide how many copies of the literal to store, because macro expansion is exactly that, a substitution of macros for their expansion that happens before the compiler takes a shot at the source code. If you want to make sure that only one copy exists, then you must write something like:
#define HELLO "Hello!"
char* hello = HELLO;
Which is equivalent to:
char* hello = "Hello!";
Also note that a declaration like:
const char* hello = "Hello!";
Keeps hello immutable, but not necessarily the memory it points to, because of:
char h = (char) hello;
h[3] = 'n';
I don't know if this case is defined in the C reference, but I would not rely on it:
char* hello = "Hello!";
char* hello2 = "Hello!"; // is it the same memory?
It is better to think of literals as unique and constant, and treat them accordingly in the code.
If you do want to modify a copy of a literal, use arrays instead of pointers, so it's guaranteed a different copy of the literal (and not an alias) is used each time:
char hello[] = "Hello!";
Back to your original question, the memory for the literal "heyapple!" will be available (will be referenceable) as long as a reference is kept to it in the running code. Keeping a whole module (a loadable library) in memory because of a literal may have consequences on overall memory use, but that's another concern (you could also force the unloading of the module that defines the literal and get all kind of strange results).
First,it IS in the code the I wrote, so it never will be take off, right? my answer is yes. I recommend you to have a look at the structure of ELF or runtime structure of executable. The position that the string literal stored is implementation dependent, in gcc, string literal is store in the .rdata segment. As the name implies, the .rdata is read-only. In your code
char *p
p = string;
the pointer p now point to an address in a readonly segment, so even after the end of function call, that address is still valid. But if you try to return a pointer point to a local variable then it is dangerous and may cause hard-to-find bugs:
int *func () {
int localVal = 100;
int *ptr = localVal;
return p;
}
int val = func ();
printf ("%d\n", val);
after the execution of func, as the stack space of func is retrieve by the c runtime, the memory address where localVal was stored will no longer guarantee to hold the original localVal value. It can be overidden by operation following the func.
Back to your question title
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string literal have static duration.
As for "And about constants like 1, 2.10, 'a'?"
my answer is NO, your can't get address of a integer literal using &1. You may be confused by the name 'integer constant', but 1,2.10,'a' is not right value ! They do not identify a memory place,thus, they don't have duration, a variable contain their value can have duration
compound literals, well, I am not sure about this.

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